Characteristic subgroups of normal subgroups are normal

Question

I know the following is already asked, but I had doubts. Let $G$ be a group, $N$ a normal subgroup of $G$, and $M$ is a characteristic subgroup of $N$. Then prove that $M$ is a normal subgroup of $G.$ I know $N$ is a normal subgroup. Therefore it is invariant under conjugation map i.e inner automorphism on $G$ i.e $i_g(N)\in N$. $M$ is a characteristic subgroup of $N$. Therefore for every automorphism on $N$ it is invariant i.e $T(M) \subset M$. But for connection between both statement I can only say now that $M$ is invariant under inner automorphism of $G$. There is no reference here to outer automorphisms that are not inner automorphisms. Where is the mistake in my argument?

Answer 1

You are not asked to prove that $M$ is invariant under any automorphism of $G$, but only under inner automorphisms.

Denote by $i_g$ the inner automorphism induced by $g\in G$. Since $N$ is normal in $G$, you know that $i_g(N)\subseteq N$.

Now prove that $i_g(N)=N$, so $i_g$ induces an automorphism of $N$, call it $\varphi$.

Since $M$ is characteristic in $N$, you know that $\varphi(M)\subseteq M$.

Therefore, $i_g(x)\in M$, for every $x\in M$. Hence $M$ is normal in $G$.